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The characteristic polynomial isTo prove triangularizability, use induction on $n$. The result is immediate for $n=1$. Over $\mathbb{C}$, ${\chi }_A$ has a root $\lambda$, so $A$ has an eigenvector $v_1$. Extend it to a basis. In this basis,By induction, a change among the remaining $n-1$ basis vectors makes $B$ upper triangular. Thus the triangularization over an algebraically closed field gives

The minimal polynomial $m_A$ is the monic polynomial of least degree satisfying $m_A(A)=0$. To establish existence without quoting the Cayley-Hamilton theorem, let $T$ be an upper-triangular matrix similar to $A$, with diagonal entries ${\lambda }_1,\dots ,{\lambda }_n$, and putThenThe factors $T-{\lambda }_{k}I$ commute, so applying all of them successively lowers the invariant flag to zero:Similarity gives the same polynomial identity for $A$. Hence a nonzero monic annihilating polynomial of degree $n$ exists, and a least-degree one exists.

If $m$ and $\tilde{m}$ were two monic annihilating polynomials of the same least degree, then $m-\tilde{m}$ would be an annihilating polynomial of smaller degree unless it were zero. Thus the minimal polynomial is unique, and the minimal polynomial bound from a triangular invariant flag gives

The eigenvalue equation $D_{r}f=\lambda f$ isIf $\lambda =-1$, this forces $f(x+r)=0$ for every $x$, hence $f=0$, so $-1$ is not an eigenvalue. If $c=1+\lambda \ne 0$, choose arbitrary values on representatives of the cosets of $r\mathbb{Z}$ and extend byThis produces nonzero eigenfunctions. Choosing functions supported on distinct cosets gives infinitely many linearly independent eigenfunctions. Thereforeand every eigenspace is infinite-dimensional.

Direct expansion giveswhich is symmetric in $r,s$. Hence

Suppose a degree-$n$ polynomial $p$ were a sum of $n$ periodic functions,Apply the commuting product $D_{r_1}\cdots D_{r_n}$. Every term on the right is killed by its corresponding factor, whereas for leading coefficient $a_n$ the mixed finite difference of a polynomial givesThis contradiction proves

Reduce coefficients modulo $I$:This is a surjective ring homomorphism and its kernel consists exactly of polynomials all of whose coefficients lie in $I$, namely $I[X]$. The first isomorphism theorem gives the coefficientwise quotient of a polynomial ringIf $I$ is prime, then $R/I$ is an integral domain, so $(R/I)[X]$ is an integral domain. The quotient criterion therefore shows that $I[X]$ is prime in $R[X]$.

Let $G={\mathbb{F}}_p^{\times }$, and let $m$ be the least common multiple of the orders of its elements. For every prime power $q^a$ dividing $m$, some element of $G$ has order divisible by $q^a$, and a suitable power of it has order exactly $q^a$. Multiplying these elements over the distinct primes produces, because their orders are coprime, an element of order $m$.

Every element of $G$ is a root of $X^m-1$. A nonzero polynomial of degree $m$ over a field has at most $m$ roots, so $p-1\le m$. On the other hand, every element order divides $p-1$ by Lagrange's theorem, so $m\le p-1$. Hence $m=p-1$, and

The squaring homomorphism has kernel $\{1,-1\}$ because $p$ is odd. Its image $H$ therefore has order $(p-1)/2$ and index two. If $p=3$, then $3=0$ is already a square. Otherwise $2,3\in G$. In the two-element quotient $G/H$, either $2H=H$, or $3H=H$, or both are the nontrivial coset, in which case $6H=H$. Thus one of $2,3,6$ is a square modulo $p$.

Finally,Whichever of $2,3,6$ is a square supplies a root, soThis is the index-two square-class argument for three related residues.

If the bound is to hold with finite $c$, the error functional must vanish on every polynomial whose fourth derivative is zero. Thus the scheme must be exact for degrees zero through three. Applying it to powers of $t=x+1$ givesSolving,This is the four-point one-sided second-derivative formula.

The Peano kernel theorem says that if a linear functional $L$ annihilates all polynomials of degree below $r$, then for $f\in C^r[a,b]$,

Here $r=4$ and $L=e$. Under the stated nonnegativity assumption,The integral equals $e(q)$ for $q(x)=(x+1{)}^4/24$, since $q^{(4)}=1$. Now $q^{\prime \prime }(-1)=0$ andThereforeEquality is attained by $q$, so the sharp Peano-kernel constant for the four-point endpoint second derivative is

The equilibrium magnetization is a global minimum of the Landau free energy. Its stationary values solve the polynomial equationand a local minimum must satisfyOne compares the value of $f$ at every such local minimum and chooses the smallest. Since $a_6>0$, the polynomial tends to positive infinity as $|m|$ tends to infinity, so a global minimum exists.

Let $E_n$ and $E$ be the projection-valued measures of $A_n$ and $A$. Using the projection $P_n:\mathcal{H}\to {\mathcal{H}}_n$, defineThe scalar spectral measures converge weakly whenfor every bounded continuous function $f$ and every $v,w\in \mathcal{H}$. By the spectral theorem for normal operators on a separable Hilbert space, this is equivalent to

The assumed moment identities say precisely that this convergence holds for every monomial $f(x)=x^m$. It follows by linearity for every polynomial. For $v=w$, the $m=2$ identity givesso Markov inequality makes the positive measures $({\mu }_v^{(n)})$ tight. Higher even moments similarly control the tails of any fixed polynomial.

Given a bounded continuous $f$ and $ϵ>0$, choose $R$ so that the measure tails are uniformly small. The Weierstrass approximation theorem supplies a polynomial $p$ withMoment convergence handles $p$; tightness and a sufficiently high even moment handle the two tails. Hence $\int fd{\mu }_v^{(n)}\to \int fd{\mu }_v$. The polarization identity then gives the same conclusion for ${\mu }_{v,w}^{(n)}$. This proves weak convergence of scalar spectral measures.

The assertion fails if only $m=1$ is assumed. Let $\mathcal{H}={\ell }^2(\mathbb{N})$, let ${\mathcal{H}}_n=\mathrm{span}\{e_1,\dots ,e_n\}$, take $A=0$, and letThe reversal matrices are self-adjoint unitaries. For fixed $v,w\in {\ell }^2$,because the finite head of one vector is paired with the vanishing tail of the other. Thus the $m=1$ condition holds. However, $A_n^2=I_{{\mathcal{H}}_n}$, soin general. Taking $f(x)=x^2$ shows that the spectral measures do not converge weakly.

Let $f$ be the common minimal polynomial. Evaluation givesso composing the two isomorphisms gives $\mathbb{Q}(\alpha )\cong \mathbb{Q}(\beta )$ and sends $\alpha$ to $\beta$.

Take $\alpha =\beta =\sqrt[4]{2}$. The field is real and contains the two roots $\pm \alpha$ of $X^4-2$, but not $\pm i\alpha$. Hence exactly two $\mathbb{Q}$-automorphisms are possible, while the degree is four.

In general, after choosing one isomorphism $\varphi :\mathbb{Q}(\alpha )\to \mathbb{Q}(\beta )$, every other one is uniquely $\varphi \circ \sigma$ for a $\mathbb{Q}$-automorphism $\sigma$ of $\mathbb{Q}(\alpha )$. ThusBy the automorphism-count divisibility theorem, this order divides $[\mathbb{Q}(\alpha ):\mathbb{Q}]$. Therefore $k$ always divides the degree.

Let $\beta$ satisfy ${\beta }^p=\alpha$. Since $K=\mathbb{Q}(\alpha )\subseteq \mathbb{Q}(\beta )$,The polynomial $f(X^p)$ has degree $p[K:\mathbb{Q}]$ and has root $\beta$. It is irreducible exactly when $[K(\beta ):K]=p$.

If $\alpha ={\gamma }^p$ for some $\gamma \in K$, then $\gamma$ is a root of $f(X^p)$ of degree at most $[K:\mathbb{Q}]$, so $f(X^p)$ is reducible. Conversely, if it is reducible, thenSince $p$ is prime, $1\le d<p$ implies $\gcd (d,p)=1$. Applying coprime-degree descent for powers to $K(\beta )/K$ and ${\beta }^p=\alpha$ shows that $\alpha$ is a $p$th power in $K$. Hence

An algebraic number is an algebraic integer when it is a root of a monic polynomial in $\mathbb{Z}[X]$.

If $\alpha$ and $\beta$ are algebraic integers, the ring $\mathbb{Z}[\alpha ,\beta ]$ is a finitely generated $\mathbb{Z}$-module. Multiplication by $\alpha \beta$ preserves this module, so the algebraic integer module criterion shows that $\alpha \beta$ is an algebraic integer.

The ring of integers of a quadratic field givessince $2\equiv 2(\mathrm{mod}4)$. For a direct verification, if $x=a+b\sqrt{2}$ with $a,b\in \mathbb{Q}$ is integral, then its trace $2a$ and norm $a^2-2b^2$ are integers. Writing $2a=m\in \mathbb{Z}$ and reducing the norm condition in lowest terms shows first that $b$ can have denominator at most $2$; a denominator $2$ with odd numerator would require $m^2\equiv 2(\mathrm{mod}4)$, which is impossible. Thus $b\in \mathbb{Z}$, and then the norm condition forces $m$ even, so $a\in \mathbb{Z}$. Conversely, every $a+b\sqrt{2}$ with $a,b\in \mathbb{Z}$ is integral because $\sqrt{2}$ is integral and algebraic integers form a ring.

Now let $f$ have degree $d$ and roots ${\alpha }_1,\dots ,{\alpha }_d$. Sinceand $|{\alpha }_1|>1$, the integer $|a_d|$ cannot be $2$; hence $|a_d|=1$. Thus $f$ is monic up to sign and ${\alpha }_1$ is an algebraic integer.

Let $B$ be the product of all roots with modulus greater than one, counting multiplicity. Complex roots occur in conjugate pairs, so $B$ is real up to the signs of the real roots andAs a product of algebraic integers, $B$ is an algebraic integer; the equality forces $B=\pm 2$. Thereforeis the absolute value of the product of the remaining conjugates, is rational, and is an algebraic integer. It is consequently a nonzero rational integer. Every remaining factor has modulus at most one, so this integer is at most one. It must equal one, and henceThis is the mahler-measure norm argument at measure two.

A nonempty topological space is irreducible when it is not the union of two proper closed subsets, equivalently when any two nonempty open subsets intersect.

The Zariski topology on ${\mathbb{A}}^n$ is Noetherian because $k[x_1,\dots ,x_n]$ satisfies the ascending chain condition on ideals. Suppose a closed set $X$ that is not a finite union of irreducible closed sets were minimal among such counterexamples. It is reducible, say $X=X_1\cup X_2$ with both $X_i$ proper closed subsets. Minimality expresses each $X_i$ as a finite union of irreducible closed sets, and combining those decompositions contradicts the choice of $X$. Thus every closed $X$ has a finite irreducible decomposition.

Using $x_3=x_2^2$, the second defining polynomial reduces toConsequentlyEach component is the image of $t\mapsto (\pm t,t,t^2)$ and is isomorphic to ${\mathbb{A}}^1$, hence irreducible. In characteristic two the two displayed components coincide.

Suppose a polynomial $P(x,y)={\sum }_{j=0}^m{p}_j(x)y^j$ vanishes on every $(x,e^x)$. Thenas an entire function. These exponential-polynomial functions are linearly independent: applying $(D-m{)}^{\mathrm{deg}{p}_m+1}$ kills the term with largest exponent, while it acts injectively on $p_j(x)e^{jx}$ for $j<m$; induction on the number of terms finishes the proof. Hence every $p_j$ is zero, so no nonzero polynomial vanishes on the graph. Its Zariski closure is all of ${\mathbb{A}}^2$.

Finally refine an open cover of an affine variety $X$ by distinguished opens $D(f_j)$, choosing one inside an original member around each point. Since these distinguished opens cover,The Hilbert Nullstellensatz implies that the $f_j$ generate the unit ideal. Thusfor finitely many of them, so $D(f_{j_1}),\dots ,D(f_{j_r})$ cover $X$. The corresponding original open sets form a finite subcover. This is quasi-compactness of an affine variety.